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A wire suspended vertically from one of ...

A wire suspended vertically from one of its ends is stretched by attaching a weight of 20 N to its lower end. If its length changes by 1% and if the Young's modulus of the material of the wire is `2 xx 10^(11) N//m^(2)`, then the area of cross section of the wire is

A

`1 mm^(2)`

B

`10^(-1) mm^(2)`

C

`10^(-2) mm^(2)`

D

`10^(-3) mm^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Y = (FL)/(Al)`
`:. A = (FL)/(lY) = (20 xx 100)/(2 xx 10^(11)) = 10^(-8) m^(2) = 10^(-2) mm^(2)`
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